By Garrett Birkhoff

This vintage, written by way of younger teachers who turned giants of their box, has formed the knowledge of recent algebra for generations of mathematicians and continues to be a necessary reference and textual content for self research and school classes.

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Qn of S such that Qi ∩ R = Pi for all 1 ≤ i ≤ n. Proof: By induction on n it suffices to prove the case n = 2. By localizing both R and S at the multiplicatively closed set R \ P2 , without loss of generality R is a local integrally closed domain with maximal ideal P2 . It suffices to prove that P1 contracts from a prime ideal contained in Q2 , or that P1 contracts from a prime ideal in the ring extension SQ2 . 4 it suffices to prove that P1 SQ2 ∩ R = P1 . 3. Lying-Over, Incomparability, Going-Up, Going-Down 33 Let r ∈ P1 SQ2 ∩ R.

Without loss of generality deg(e1 ) > · · · > deg(en ). If deg(e1 ) > 0, the homogeneous part of e2 − e of degree 2 deg(e1 ) is exactly e21 , and since e2 − e = 0 and R is reduced, we get a contradiction. Similarly if deg(en ) < 0 we get a contradiction. This forces n = 1 and e to be homogeneous of degree 0, whence proving (3). Assume (3). To prove (4), by symmetry it suffices to prove (4) for i = 1. Consider the idempotent e = (1, 0, . . 13). By assumption e is homogeneous. 3. Integral closure and grading 37 elements a, b ∈ R, with b a non-zerodivisor in R such that e = a/b.

If (F1 , . . , Fm ) = R1 R, then (F1 , . . , Fm ) = R1 R. Proof: By assumption there exists n such that R1n ⊆ (F1 , . . , Fm ). By homogeneity, R1n R ⊆ R1n−1 (F1 , . . , Fm ). This proves that (F1 , . . 5 that R1 R ⊆ (F1 , . . , Fm ). But R1 R is a radical ideal, so R1 R ⊆ (F1 , . . , Fm ) ⊆ (F1 , . . , Fm ) ⊆ R1 R, which finishes the proof. , µ(m) > dim R, and let I be an ideal of finite projective dimension. Then m(I : m) = mI and I : m is integral over I. Proof: Without loss of generality I = R.